From Lagrangian to Hamiltonian canonical equations

[Leo32]

Hi y'all,

 

Working myself at walking pace through Shankar's "Principles of Quantum Mechanics", there is some intreguing jump towards Hamiltons canonical equations that I'm missing.

It could be something obvious, since Wikipedia doesn't provide any additional info either.

 

So far, the canonical momentum in Lagrangian formalism was defined as

p= dL/dq(dotted)

 

Terribly sorry, but I can't get the LaTex working... and I know that should be partial derivates, and also imagine the p and q's being indexed.

 

A certain deduction, which I still understand results in

dH/dq = -dL/dq

 

The strange thing, which I don't understand at all, is that dL/dq is equalled to p(dotted).

 

No idea where this pops out of, or what the reasoning is behind it. I see it provides a nice symmetry in Hamilton's canonical equations, but other than that, I see no reason why.

Given that p, q and q(dotted) are defined yet, it's not just a question of defining a new thing called p(dotted) I suppose, so where oh where is the reason behind this ?

 

Cheers !

Leo

[□h=-16πT]

Lagrangian function:

 

L=T(x',y',z') - V(x,y,z) [primes denote time derivatives]

 

Only the kinetic energy is an explicit function of the time derivatives and so partial differentiation of the lagrangian with respect to these is the derivative of the kinetic energy. The kinetic energy of a body under the influence of a conservative force is quadratic in the time derivatives and so upon differentiation we end up with the momentum of the system. For example in Cartesian coordinates

 

1/2(mx')^2 -differentiate with respect to x'-> mx'

 

which is the equation for momentum. Using a few more examples of coordinates or just a little bit of quick thought it's simple to see how this can be generalised.

[timo]

The strange thing' date=' which I don't understand at all, is that dL/dq is equalled to p(dotted).

[/quote']

That´s no shame if you can´t understand why it is like that. Reason: That´s simply the definition of the canonical momentum. Of course, it´s chosen in a way that it fits the momentum you know from Newtonian physics. But it still is simply a definition and not a deduction from something else.

 

EDIT: And you can´t get LaTeX working because it´s not working atm so that´s no fault of yours.